Repository Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2932    Accepted Submission(s): 1116 Problem Description When you go shopping, you can search in repository for avalible merchandises…

题目 字典树,注意初始化的位置~!!位置放错,永远也到不了终点了org.... 我是用数组模拟的字典树,这就要注意内存开多少了,,要开的不大不小刚刚好真的不容易啊.... 我用了val来标记是否是同一个串分解而来的,保存的是串的编号 num记录数目. //string &replace(iterator first0, iterator last0,const_iterator first, const_iterator last); //把[first0,last0)之间的部分替换成[firs…

RepositoryTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 7233    Accepted Submission(s): 2278 Problem DescriptionWhen you go shopping, you can search in repository for avalible merchandises by…

Repository Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 5048    Accepted Submission(s): 1739 Problem Description When you go shopping, you can search in repository for avalible merchandises b…

Repository Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 6444    Accepted Submission(s): 2096 Problem Description When you go shopping, you can search in repository for avalible merchandises b…

字典树较为复杂的应用,我们在建立字典树的过程中需要把所有的前缀都加进去,还需要加一个id,判断它原先是属于哪个串的.有人说是AC自动机的简化,但是AC自动机我还没有做过. #include<iostream> #include<string.h> #include<cstdio> using namespace std; ],b[]; struct Node { int id,num; Node *sons[]; Node() { id = -,num = ; ; j…

http://acm.hdu.edu.cn/showproblem.php?pid=2846 Repository Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2720    Accepted Submission(s): 1060 Problem Description When you go shopping, you can s…

http://acm.hdu.edu.cn/showproblem.php?pid=1979 Fill the blanks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 373    Accepted Submission(s): 155 Problem Description There is a matrix of 4*4, yo…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1671 Problem Description Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers: 1. Emergenc…

把每个字符串的所有子串都加入字典树,但在加入时应该注意同一个字符串的相同子串只加一次,因此可以给字典树的每个节点做个记号flag--表示最后这个前缀是属于那个字符串,如果当前加入的串与它相同,且二者属于不同串就自加,否则不作处理. AC代码: #include<cstdio> #include<cstring> const int Max=26; struct node{ node *next[Max]; int cnt,num; //用num记录最后一次加入的字符串编号,若不同串…